The Natural Exponential Function e^x — Edexcel A-level Mathematics (Pure)

SPEC 6.2 • EXPONENTIALS AND LOGARITHMS

The Natural Exponential Function eˣ

Understanding Euler's number and the unique properties of the natural exponential

What is e?

Euler's Number

e is an irrational number approximately equal to:e ≈ 2.71828182845...

Like π, the number e is one of the most important constants in mathematics. It appears naturally in:

Compound interest calculations

Population growth models

Radioactive decay

Calculus and differential equations

Calculator

On your Casio calculator, press SHIFT then ln to access eˣ. Or use the e button directly.

The value of e

The number e can be defined in several equivalent ways:

Definition 1: Limit

e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}

Definition 2: Series

e = 1 + \frac{1}{1 !} + \frac{1}{2 !} + \frac{1}{3 !} + \frac{1}{4 !} + ...

Key Values

ExpressionValuee⁰1e¹2.718... (≈ 2.72)e²7.389... (≈ 7.39)e⁻¹0.368... (≈ 0.37)

Why is e special?

The number e is the only base where the gradient of the exponential function equals the function itself.

The Unique Property

If y = e^{x}, then \frac{d y}{d x} = e^{x}

The gradient at any point equals the y-coordinate at that point!

At x = 0

y = e⁰ = 1

Gradient = e⁰ = 1

At x = 1

y = e¹ ≈ 2.718

Gradient = e¹ ≈ 2.718

Why "Natural"?

This unique gradient property is why y = eˣ is called the natural exponential function — it's the most natural choice of base for calculus.

Differentiating eˣ

Key Result

\frac{d}{d x} (e^{x}) = e^{x}

This result extends to functions of the form eᵏˣ:

Chain Rule Result

\frac{d}{d x} (e^{k x}) = k e^{k x}

Function yDerivative dy/dx

eˣeˣ

e²ˣ2e²ˣ

e⁻ˣ−e⁻ˣ

e³ˣ3e³ˣ

3eˣ3eˣ

Pattern

Differentiate eᵏˣ by multiplying by k. The exponential part stays the same!

Graph of y = eˣ and its derivative

Key Observation

The curve y = eˣ and its derivative dy/dx = eˣ are exactly the same curve! The purple dashed line sits directly on top of the green solid line. This is the unique property of base e.

01Worked Example

Differentiate the following:

(a) y = e⁴ˣ (b) y = 5e⁻²ˣ (c) y = e³ˣ + eˣ

(a) y = e⁴ˣ

Using d/dx(eᵏˣ) = keᵏˣ with k = 4:

dy/dx = 4e⁴ˣ

(b) y = 5e⁻²ˣ

Using d/dx(eᵏˣ) = keᵏˣ with k = −2:

dy/dx = 5 × (−2)e⁻²ˣ = −10e⁻²ˣ

(c) y = e³ˣ + eˣ

Differentiate each term:

dy/dx = 3e³ˣ + eˣ = 3e³ˣ + eˣ

Worked Example02

Find the gradient of y = eˣ at the point where x = 2.

Step 1: Differentiate

y = eˣ

dy/dx = eˣ

Step 2: Substitute x = 2

Gradient = e² = 7.39 (3 s.f.)

Check: When x = 2, y = e² ≈ 7.39

The gradient equals the y-coordinate — as expected!

Worked Example03

Sketch the graph of y = eˣ + 3, stating the y-intercept and the equation of the asymptote.

Step 1: Identify the transformation

y = eˣ + 3 is y = eˣ translated 3 units up

Step 2: Find the new asymptote

Original asymptote: y = 0

New asymptote: y = 0 + 3 = y = 3

Step 3: Find the y-intercept

When x = 0: y = e⁰ + 3 = 1 + 3 = 4

Y-intercept: (0, 4)

Step 4: Sketch the curve

Same shape as y = eˣ, but shifted up 3 units.

Worked Example04

On the same axes, sketch y = eˣ and y = e⁻ˣ. State the coordinates where they intersect.

Sketching:

y = eˣ: exponential growth through (0, 1)

y = e⁻ˣ: exponential decay through (0, 1) — reflection in y-axis

Finding intersection:

Set eˣ = e⁻ˣ

This means x = −x (comparing exponents)

2x = 0, so x = 0

When x = 0: y = e⁰ = 1

Intersection point: (0, 1)

05Worked Example

Sketch the graph of y = 2eˣ⁻¹ − 1, stating:

(a) the y-intercept (b) the asymptote (c) the transformations from y = eˣ

(c) Transformations from y = eˣ:

1. Horizontal translation 1 unit RIGHT (the x−1)

2. Vertical stretch by factor 2 (the coefficient 2)

3. Vertical translation 1 unit DOWN (the −1)

(b) Asymptote:

Only vertical translation affects asymptote

Asymptote: y = 0 − 1 = y = −1

(a) Y-intercept:

When x = 0: y = 2e⁰⁻¹ − 1 = 2e⁻¹ − 1

y = 2(0.368) − 1 = 0.736 − 1 = −0.264

Y-intercept: (0, 2e⁻¹ − 1) ≈ (0, −0.26)

Worked Example06

Find the equation of the tangent to y = eˣ at the point (1, e). Give your answer in the form y = mx + c.

Step 1: Differentiate to find gradient function

y = eˣ → dy/dx = eˣ

Step 2: Find gradient at x = 1

m = e¹ = e

Step 3: Use y − y₁ = m(x − x₁)

Point: (1, e), gradient: m = e

y − e = e(x − 1)

y − e = ex − e

y = ex

Answer: y = ex (or y = ex + 0)

Practice

Question 1: Differentiate y = e⁵ˣ − 2e⁻ˣ

Differentiate each term:

d/dx(e⁵ˣ) = 5e⁵ˣ

d/dx(−2e⁻ˣ) = −2 × (−1)e⁻ˣ = 2e⁻ˣ

dy/dx = 5e⁵ˣ + 2e⁻ˣ

Practice

Question 2: Find the gradient of y = e²ˣ at the point where x = 1.

Step 1: Differentiate

y = e²ˣ

dy/dx = 2e²ˣ

Step 2: Substitute x = 1

Gradient = 2e²⁽¹⁾ = 2e²

= 2 × 7.389...

= 14.8 (3 s.f.)

Practice

Question 3: Sketch y = eˣ − 2, stating the asymptote and y-intercept. Does the curve cross the x-axis?

Asymptote: y = 0 − 2 = y = −2

Y-intercept: When x = 0, y = e⁰ − 2 = 1 − 2 = −1

Y-intercept: (0, −1)

X-axis crossing: Set y = 0

eˣ − 2 = 0

eˣ = 2

x = ln 2 ≈ 0.693

Yes, it crosses at (ln 2, 0)

Summary

Key Results

ConceptKey Information

Euler's number ee ≈ 2.71828... (irrational constant)

Derivative of eˣd/dx(eˣ) = eˣ

Derivative of eᵏˣd/dx(eᵏˣ) = keᵏˣ

Graph of y = eˣ(0, 1), asymptote y = 0, range y > 0

Unique propertyGradient at any point = y-coordinate

Exam Essentials

d/dx(eˣ) = eˣ — the function equals its own derivative

d/dx(eᵏˣ) = keᵏˣ — multiply by the coefficient of x

The gradient of y = eˣ at any point equals the y-coordinate

y = e⁻ˣ is the reflection of y = eˣ in the y-axis